\( \mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}\) From the reaction above, it can be deduced that ____________
A. \(\mathrm{Zn}\) is reduced B. \(\mathrm{Zn}\) is the oxidizing agent C. Cu loses electrons oxidizing agent D. \(\mathrm{Cu}\) is the oxidizing agent
Correct Answer: D
Explanation
\(\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}\) \(\Rightarrow\) Since the oxidation number of \(\mathrm{Zn}\) increases from 0 to \(+2, \mathrm{Zn}\) has been oxidizeD. It is therefore, the reducing agent. \(\Rightarrow\) Since the oxidation number of \(\mathrm{Cu}^{2+}\) decreases from \(+2\) to \(0, \mathrm{Cu}^{2+}\) has been reduced; it is therefore the oxidizing agent.
